2021 Fall AMC 12A Problems/Problem 22

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Problem

Azar and Carl play a game of tic-tac-toe. Azar places an in $X$ one of the boxes in a 3-by-3 array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all boxes are filled or one of the players has of their symbols in a row—horizontal, vertical, or diagonal—whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O$. How many ways can the board look after the game is over?

$\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160$

Solution

There are six ways for Carl to win with a complete row or column. For each way, there are six spaces where Carl did not play and Azar could have played. So, there are ${6 \choose 3}=20$ ways that Azar could have played. However, two of these ways would lead to Azar winning, so they must be excluded. This leads to $6(20-2)=108$ ways the board could look after the game is over.

Also, there are two ways for Carl to win with a complete diagonal. For each way, there are six spaces and ${6 \choose 3}=20$ ways Azar could have played. This leads to $2\cdot20=40$ ways the board could look.

In total, there are $108+40 \implies \boxed{\textbf{(D) } 148}$ ways the board could look.

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 21
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